A particle of mass `m` experienced an elastic collision with a stationary particle of mass `2m` . What fraction of the kinetic energy does the striking particle lose if recoils at the right angles to its original motion direction.

Explanation

Since the total momentum before the collision is along the `x`-axis , the total momentum after the collision should be along the `x`-axis, for this particle of mass `2m` should make at angle so that the net momentum along the `y`-axis is zero.
Fractional loss of `K.E.` of striking particle is
`((1)/(2) m u^(2) - (1)/(2) mv_(1)^(2))/((1)/(2) m u^(2)) = 1 - ((v_(1))/(u))^(2)`
By the momentum conservation:
Along the `x`-axis:
`mu = 2m v_(2) cos theta rArr v_(2) cos theta = (u)/(2)` (i)
Along the `y`-axis:
`0 = mv_(1) - 2mv_(2) sin theta rArr v_(2) sin theta = (v_(1))/(2)` (ii)
Eliminating `theta` from (i) and (ii) , we get
`(v_(2) cos theta)^(2) + (v_(2) sin theta)^(2) = (u^(2) + v_(1)^(2))/(4)`
`v_(2)^(2) = (u^(2) + v_(1)^(2))/(4)`
Since the collision is elastic
`K_(i) = K_(f)`
`(1)/(2) m u^(2) = (1)/(2) mv_(1)^(2) + (1)/(2) xx 2m v_(2)^(2)`
`u^(2) = v_(1)^(2) + 2v_(2)^(2) = v_(1)^(2) + 2((u^(2) + v_(1)^(2))/(4)) rArr ((v_(1))/(u))^(2) = (1)/(3)`
Fractional loss `= 1 - ((v_(1))/(u))^(2) = 1 - ((1)/(3)) = (2)/(3)`